Sometimes some words like “localization” or “internationalization” are so long that writing them many times in one text is quite tiresome.
Let’s consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn’t contain any leading zeroes.
Thus, “localization” will be spelt as “l10n”, and “internationalization» will be spelt as “i18n”.
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
The first line contains an integer n (1 ≤ n ≤ 100). Each of the following n lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
第一行包含了一個整數 n (1 ≤ n ≤ 100)。 接下來的 n 行都包含了一個單字。每個單字是由小寫的字母組成,且長度會在 1 到 100之間。
輸出
Print n lines. The i-th line should contain the result of replacing of the i-th word from the input data.
印出 n 行。 ith 行應該包含輸入資料 ith 的單字替換後的結果。
範例
Input
4 word localization internationalization pneumonoultramicroscopicsilicovolcanoconiosis
Output
word l10n i18n p43s
解題思路
這一題其實要考得很簡單
就是將長度超過10的單字,留頭留尾,中間則是其餘字元的數量
所以我們就是要先判斷初始長度,就可以輕鬆解決這個問題了
C#解決方案
方案1
int n = int.Parse(Console.ReadLine());
for(int i=0;i<n;i++){
string str = Console.ReadLine();
if(str.Length>10){
Console.WriteLine(str[0]+(str.Length-2).ToString()+str[str.Length-1]);
} else {
Console.WriteLine(str);
}
}
Java解決方案
方案1
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = Integer.parseInt(scanner.nextLine());
for (int i = 0; i < n; i++)
{
String str = scanner.nextLine();
if(str.length()>10){
System.out.println(str.charAt(0)+String.valueOf(str.length()-2)+str.charAt(str.length()-1));
} else {
System.out.println(str);
}
}
}
}
Python3解決方案
方案1
n = int(input())
for i in range(n):
x = input()
if (len(x) > 10):
print(x[0] + str(len(x)-2) + x[len(x)-1])
else:
print(x)
JavaScript解決方案
方案1
var n = readline();
for (var i = 0; i < n; i++) {
var str = readline();
if (str.length > 10) {
print(str[0] + (str.length - 2) + str[str.length - 1]);
} else {
print(str);
}
}
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